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Step-by-step Solution
Step 1: Identify the Reaction and Order
The reaction under consideration is the decomposition of hydrogen peroxide:
$H_2 O_2 (aq) \rightarrow H_2 O (aq) + \tfrac{1}{2}\,O_2(g)$
It is given that this decomposition follows a first-order reaction.
Step 2: Calculate the Rate Constant, k
For a first-order reaction, the rate constant k can be found using the formula:
$k = \dfrac{2.303}{t} \,\log \,\biggl(\dfrac{a}{a-x}\biggr)$
Here,
$a =$ initial concentration of $H_2 O_2 = 0.5\,\text{M}$
$(a-x) =$ final concentration of $H_2 O_2 = 0.125\,\text{M}$
$t = 50\,\text{minutes}$
Substitute the values:
$k = \dfrac{2.303}{50} \,\log \biggl(\dfrac{0.5}{0.125}\biggr)$
Calculating the logarithm:
$\log \biggl(\dfrac{0.5}{0.125}\biggr) = \log (4) \approx 0.6021
Hence,
$k \approx \tfrac{2.303}{50} \times 0.6021 \approx 2.78 \times 10^{-2}\,\text{min}^{-1}$
Step 3: Determine the Rate of Decomposition at $[H_2 O_2] = 0.05\,\text{M}$
The rate of a first-order reaction is given by:
$r = k\,[H_2 O_2]$
Substituting $k = 2.78 \times 10^{-2}\,\text{min}^{-1}$ and $[H_2 O_2] = 0.05\,\text{M}$ gives:
$r = (2.78 \times 10^{-2}) \times (0.05) = 1.386 \times 10^{-3}\,\text{mol min}^{-1}$
This is the rate of decomposition of $H_2 O_2$ (i.e. the rate at which $H_2 O_2$ is consumed).
Step 4: Relate the Rate of Decomposition to the Rate of Formation of $O_2$
From the stoichiometry of the reaction:
$-\,\dfrac{d[H_2 O_2]}{dt} \;=\; \dfrac{2\,d[O_2]}{dt}
Rearranging to solve for the rate of formation of $O_2$:
$\dfrac{d[O_2]}{dt} \;=\; \dfrac{1}{2}\,\Bigl(\!-\dfrac{d[H_2 O_2]}{dt}\Bigr)
Since $-\dfrac{d[H_2 O_2]}{dt} = 1.386 \times 10^{-3}\,\text{mol min}^{-1}$, we get:
$\dfrac{d[O_2]}{dt} = \dfrac{1}{2}\, (1.386 \times 10^{-3}) = 6.93 \times 10^{-4}\,\text{mol min}^{-1}
This is the rate of formation of $O_2$ at the time when the concentration of $H_2 O_2$ is $0.05\,\text{M}$.
Step 5: Final Answer
The rate of formation of $O_2$ is $6.93 \times 10^{-4}\,\text{mol min}^{-1}$.
